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ON THE DIOPHANTINE EQUATION x3+y3+z3=++

A. OPPENHEIM 1. The remarks in this note on the Diophantine equation

(1) x3 + yl + 33 = x + y + z

are prompted by Edgar' recent note []. In order "to avoid certain trivial solutions" he assumes that x^y S;0, z<0 and x^ —y. Using a method of S. D. Chowla and others (a reference not accessible to me) he obtains infinitely many solutions of (1) subject to the further con- ditions (2) x + y + z = , x + y = km, x + z & 0 in each of the following cases (i) = 3 (Chowla), (ii) fe= 12 (Edgar), (iii) £ = 16/3 (Edgar). In this note I show that each of the trivial solutions Qi, 1, —) where \h\ =g2, gives rise to infinitely many nontrivial solutions and that nontrivial solutions likewise generate others. As an example the equation

(3) N2 - 85M2 = - 4 has infinitely many integral solutions (A, M), both odd or both even. The integers (4) x = \iM + ), y = \iM - N), z = - 41f will always satisfy (1). And for those solutions x + y + z = — 3M, x + y = M. These solutions were obtained in fact by the method below from the nontrivial solution (5, —4, —4). The equation (5) 3A2 - 31Jkf2 = - 4 also has infinitely many solutions with M, N oi like parity: the equa- tions x + y = — M, x — y = N, z = 2M will yield solutions of (1). Equation (5) was derived from the trivial solution (-2, 1, 2) of (1). The equation Received by the editors May 19, 1965. 493 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 494 A. OPPENHEIM [April

(6) 5A2 - 62JI72 = 2 has infinitely many solutions in integers A(even) and M: the smallest solution appears to be (412, 117). Determine x, y, z by the equations x + y = 10A7, x — y = N, z = — 7J17: then x, y, z are integers which satisfy (1). One solution is therefore x = 791,y = 379, z=-819. 2. Suppose that (x, y, z) is a solution of (1). Any permutation yields another solution (not necessarily distinct). Also (—x, —y, —s) is a solution. In general 12 solutions arise from a given solution. Suppose that x+y and z are not both zero. Define integers m, re, a, uniquely by the following equations (7) x Ar y = am, z = — cm, x — y = n, (a, c) = 1, = 1. Then from the identity 4(x3 A- y3 A- z3 — x — y — z) = m{3an2 + (a3 — Ac3)m2 — i(a — c)}

we see that the integers (m, re) (m ^ 1) satisfy the Diophantine equa- tion (8) (a3 - 4c3)M2 Ar 3aN2 = 4(a - c). Conversely, suppose that integers a and c exist such that (8) is solvable in integers M9*0, N with aM, N of like parity; then the equations X + = aM, Z = - cM, X - Y = N give integers X, Y, Z which satisfy (1). If in addition the integer defined by (9) D = 3a(1c3 - a3)

is positive and not a square, then the equation (8), having one solu- tion (M, N) with aM, N of same parity, will have infinitely many such solutions, by a classical theorem on indefinite binary quadratic forms. As an example, take the trivial solution (x, y, z) = (h, 1, -h)

where h is an integer, \h\ =2. Equation (8) becomes 3(h A- l)N2 - (3/z3 - 3h2 - 3h - 1)M2 = 4. (10) D = 3(k + l)(3h3 - 3h2 -3h- 1).

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1966] THE DIOPHANTINE EQUATIONx*+?+s*°~x+y+z 495

It is easily seen that P>0 and that D is not a square whenever \h\ ^2. Now (10) has the solution M=l, N = h —l: it has therefore infinitely many solutions such that ih + l) M, N have the same par- ity. I omit the proof. Here are a few examples of (10): 9A2 - 5M2 = 4, 31Af2 - 3A2 = 4, 3N2 - 11M2 = 1, 100M2 - 6N2 = 4, 15A2 - 131M2 = 4, 229M2 - 9N2 = 4, 18xV2- 284M2 = 4, 109M2 - 3A2 = 1.

3. Each solution (x, y, z) of (1) gives rise in general to three pairs of integers (a, c) and hence to three binary forms. Suppose (, yi, zi) and (x2, y2, z2) are two solutions of (1) derived from two pairs (Ai, M/), (A2, M2), belonging to a particular binary form corresponding to a pair (a, c): suppose also that (xi, yi, zi)^(x2, y2, z2) or to (—x2, —y2, —z2). Then the two remaining binary forms deducible from the triad (xi, yi, Zi) are distinct from those deducible from the triad (x2, y2, z2). In this way further sets of solutions can be generated. As an example the trivial solution (2, 1, —2) leads to the form 9N2 —5M2 = i. The solution (6, 8) of the latter equation gives the triad (15, 9, —16) which satisfies (1). This triad yields the triads (15, —16, 9), (9, —16, 15) whence we get two sets for a, c, m, ra and two forms: -1, -9,1,31: 2915M2 - 3N2 = 32; -7, -15, 1, 25: 13157M2 - 21A2 = 32.

From these two binary forms infinitely many others can be gener- ated, each of which will lead to solutions of (1). 4. Edgar [l] gives a solution of (1) corresponding in his notation to £ = 16/3; in my notation a = 16, c=13. The corresponding equation (8) is 4A2 - 391M2 = 1 which has (as Edgar says) infinitely many solutions with even N, the smallest solution yielding x = & + , y = 8u — v, z = 13m where u —371133, v = 1834670. In the way described further forms can be generated from the permutations (8m + v, —13m, 8m — v), (8m — v, —I3u, 8m + v).

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 496 A. OPPENHEIM

The pair (a, c) = (10, 7) is also worthy of note: it leads to the form (6) which gives rise to infinitely many solutions of (1). Two more examples can be given of small a, c: a = 14, c = 11; a = 64, c = 61. The first leads to the equation 7A2 - 430AP = 2, solvable infinitely often with N even, .. Af= 2124, il7 = 271 whence x = 2959, y = 835, z = - 2981. The second (64, 61) leads to the equation (4A)2 - 53815M2 = 1 which is in fact solvably infinitely often with N even so that solutions of (1) are given by x + y = 6iM, x — y = N, z = — 61M. 5. A difficult problem remains for consideration. Two solutions of (1) may be regarded as dependent if they can be connected by a finite number of binary forms as described above. Can simple criteria be determined for dependence? Can the independent solutions be completely specified? I am grateful to Professor E. S. Barnes for some helpful comments. Reference 1. H. M. Edgar, Some remarks on the Diophantine equation x3+y3+z3 = x+y+z, Proc. Amer. Math. Soc. 16 (1965), 148-153. University of Malaya, Kuala Lumpur, Malaysia

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